### Video Transcript

A bullet is shot horizontally with
a speed of 2.0 times 10 to the two meters per second from a height of 1.5
meters. How much time elapses before the
bullet hits the ground? How far does the bullet travel
horizontally?

As we work this exercise, we’ll
assume that 𝑔, the acceleration due to gravity, is exactly 9.8 meters per second
squared. We’re told that the bullet is shot
with an initial speed of 2.0 times 10 to the two meters per second. Let’s call that speed 𝑣 sub 𝑥,
for its speed in the 𝑥 or horizontal direction. The bullet’s initial height is
given as 1.5 meters. We’ll call that value ℎ.

The problem involves two parts. For the first part, we wanna solve
for the time elapsed before the bullet hits the ground. We’ll call that time 𝑡. In part two, we wanna solve for the
horizontal displacement of the bullet. We’ll call that 𝑑 sub 𝑥.

Let’s begin our solution by drawing
a diagram of the situation. In our scenario, the bullet is
fired from the gun with an initial speed 𝑣 sub 𝑥. At first, the bullet is moving only
horizontally. But then, the force of gravity
begins to affect the path of the bullet’s flight. Ultimately, the bullet travels a
horizontal distance we’ve called 𝑑 sub 𝑥. Before solving for that distance,
we want to know how long it takes for the bullet to reach the ground. Since the acceleration acting on
the bullet is constant, that means we can use then kinematic equations to approach
this problem. Looking over the equations of
motion that are true when acceleration 𝑎 is constant, we see that the third one
from the top can be useful to us to solve for the time 𝑡.

Before we put this equation to use,
looking back at our diagram, let’s define vertical downward motion as positive and
horizontal motion to the right as positive. Under this definition, 𝑔 is equal
to positive 9.8 meters per second squared. As we apply this kinematic equation
to our scenario, we focus on motion exclusively in the vertical or 𝑦-direction. In this direction, 𝑑, the distance
the bullet travels, is ℎ. 𝑣 zero, the initial speed of the
bullet in the vertical direction, is zero, so that entire term goes to zero. Our acceleration 𝑎 is equal to 𝑔,
the acceleration due to gravity, and 𝑡 is the time it takes for the bullet to fall
to the ground. So our revised equation says that
ℎ, the vertical height for which the bullet drops, is equal to one-half 𝑔 times 𝑡
squared. Let’s rearrange this equation to
solve for 𝑡.

We’ll start by multiplying both
sides by two divided by 𝑔, which cancels out the one-half with the two and both
factors of 𝑔 on the right-hand side. If we then take the square root of
both sides, that square root cancels out with the squared term leaving us with an
equation for 𝑡 that says: 𝑡 is equal to the square root of two times ℎ divided by
𝑔. When we insert the given values for
ℎ and the known value for 𝑔 and enter these values on our calculator, we find a
value for the time 𝑡 of 0.55 seconds. That’s how long it takes the fire
bullet to fall to the ground.

Now we turn our attention to
solving for 𝑑 sub 𝑥, the horizontal range of the bullet. In the horizontal direction, the
bullet moves with a constant speed. There is no acceleration. When speed is constant, the
relationship between speed, distance, and time is given as: speed is equal to
distance travelled divided by time. In our situation, 𝑣 sub 𝑥 is
equal to 𝑑 sub 𝑥 divided by 𝑡. To solve for 𝑑 sub 𝑥, let’s
multiply both sides by 𝑡 which cancels that term from the right-hand side. So 𝑑 sub 𝑥 is equal to 𝑡 times
𝑣 sub 𝑥. We solved for the value of 𝑡
previously and we’re given 𝑣 sub 𝑥 in the problem statement.

When we insert those two values and
multiply them together, we find a distance 𝑑 sub 𝑥, to two significant figures, of
110 meters. That’s the horizontal distance the
bullet will travel before it hits the ground.